## Intro Analysis question

For the discussion of math. Duh.

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skt2k4
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### Intro Analysis question

I'm trying to teach myself Analysis to see if I can cut a math major. In troubling news, proving this elementary theorem threw me for a spin.

f(A n B) =/= f(A) n f (B)

The image of the intersection of sets A and B does not necessarily equal the the intersection of the images of A and B.

SimonM
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### Re: Intro Analysis question

I'm guessing you have to find a function which acts as a contradiction?

In which case, consider A = R-, B = R+, f(x) = x^2
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Token
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### Re: Intro Analysis question

What you could do is prove an inclusion instead of an equality.
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skt2k4
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### Re: Intro Analysis question

Here's what I came up with, which I submit with the humility of someone swimming in new, untested waters.

y is a mmeber of the set f(A) n f(B). Suppose also that y does not correspond to a single x in both sets A and B, but rather to a c in A and a d in B. In this circumstance, although y is a member of the set f(A) n f(B), its preimage is not a member of the set A n B.

Also, is this written in proper form for a proof?

skt2k4
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### Re: Intro Analysis question

I forgot to add, thanks for the quick replies. I really appreciate the help.

antonfire
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### Re: Intro Analysis question

Your proof has some problems with it. In particular, you assume in a few places that the preimage of an element is an element, when in fact it is a set. Also, you need to explicitly show that the situation you described ("y does not correspond to a single x in both sets A and B, but rather to a c in A and a d in B") is possible.

The usual way to prove a statement like this is to come up with an explicit example where f(A n B) is not f(A) n f(B).

Edit: also, it's not a big deal, but try to use the edit button instead of double posting.
Last edited by antonfire on Sun Jul 06, 2008 8:44 pm UTC, edited 1 time in total.
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ramblingsimon
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### Re: Intro Analysis question

I took Intro to Analysis last semester as a Freshman. I hated that class so much. Got a C+ and I remember nothing from it.

Good luck to you, though!
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z4lis
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### Re: Intro Analysis question

skt2k4 wrote:Here's what I came up with, which I submit with the humility of someone swimming in new, untested waters.

y is a mmeber of the set f(A) n f(B). Suppose also that y does not correspond to a single x in both sets A and B, but rather to a c in A and a d in B. In this circumstance, although y is a member of the set f(A) n f(B), its preimage is not a member of the set A n B.

Also, is this written in proper form for a proof?

You've got the right idea. The above poster might seem silly, but keep in mind that the preimage (or inverse image) is *not* a single element. Images and preimages are *sets*. That one threw me for a nasty loop until I realized that unless your function is a bijection, f-1(y) is a set, even if that set contains only a single element. {x}, for example. There is a subtle difference between just an x and {x} that I did not pick up immediately. Likewise, there is a subtle difference between f-1(y) and f-1({y}). The first is the value of the inverse function of f, and is not defined and cannot be used unless f is a bijection. f-1(y) yields non-set values (unless your function's domain is actually a set of sets). On the other hand, f-1({y}) is the inverse image of {y}. It is always defined, regardless of whether or not f is a bijection, and its values are always sets, even if that set contains only a single element.

Anyway, enough with that little explanation! To make this proof, the best way is above, by contradiction. Suppose "for all functions f:X->Y, f(A)nf(B) = f(AnB), where A and B are subsets of X." Just take a function that isn't 1-1 (the equality holds for functions that are 1-1), such as x2 and then pick two sets that lie on either side of the "bend". Maybe [-2,-1] and [1,2]. (Using interval notation, here)

Say A = [-2,-1] and B = [1,2]. Obviously, f(A) = [1,4] and f(B) = [1,4]. However, f(AnB) = f(null) = null and f(A)nf(B) = [1,4]. We have a problem with our above statement, which means it must be false.

Trying to do a proof without contradiction, which is what you did, would have to have something like "suppose f is not 1-1 ==> there exist a!=b in X such that f(a) = f(b). a!=b ==> {a}n{b} = null ==> f({a}n{b}) = null. f({a})=f({b}) (this bit requires a statement on the uniqueness of function values, bleh) ==> {f({b})}n{f({a})} = {f({a})}n{f({a})} = {f({a})}, but {f({a})} != null ==> f(A)nf(B) != f(AnB), letting A={a} and B={b}" (Granted, some quantifies are missing from that bit, I think... and moving from a!=b to {a}n{b} = null isn't too rigorous, but... phooey ) Notice that it's a bit harder, and the exact same thing is accomplished with the much more intuitive counterexample

Proving the equality for a function that's 1-1 is a probably a good exercise.
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Heahengel
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### Re: Intro Analysis question

I don't know if this will help, but I'll give the counterexample that flew into my head as I read the question.

Define f:AuB -> Z such that if a is in A\B or B\A, f(a)=1 and if a is in AnB, f(a)=0.

Then f(AnB)= {0} and f(A) n f(B)= {0,1}, so long as A\(AnB) and B\(AnB) are nonempty (neither set is contained in the other).

If you haven't had much experience with proof based math courses, I wouldn't worry that you couldn't prove this off the bat. It takes time and practice to build up a sense of how to approach problems. When I read 'prove that this is not always true,' I see 'provide a counterexample.'

Edit (for additional accuracy): The above counterexample gives a contradiction with the stated requirements, but AnB needs to be nonempty also to produce those sets. I guess its simpler to take the function g such that g(a)=0 for all a, and then require A and B to be disjoint.

Tac-Tics
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### Re: Intro Analysis question

Code: Select all

f(0) = 1f(1) = 1f({0} n {1}) = f({}) = {}f({0}) n f({1}) = {1} n {1} = {1}

Yakk
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### Re: Intro Analysis question

Tac-Tics wrote:

Code: Select all

f(0) = 1f(1) = 1f({0} n {1}) = f({}) = {}f({0}) n f({1}) = {1} n {1} = {1}

Even be nicer (some people don't like the empty set):

f(0) = 1
f(1/2) = 50
f(1) = 1

f({0, 1/2} n {1, 1/2}) = f({1/2}) = {50}
f({0,1/2}) n f({1,1/2}) = {1,50} n {1,50} = {1,50}

You need a 2-1 (or more -- ie, non-injective) component to your function for that equation to not-work. That might help you understand what is going on.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Tac-Tics
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### Re: Intro Analysis question

Yakk wrote:some people don't like the empty set):

Some people should probably learn to like it if they're going to keep reading threads on analysis =-/

emo1truth
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### Re: Intro Analysis question

Yakk wrote:
You need a 2-1 (or more -- ie, non-injective) component to your function for that equation to not-work. That might help you understand what is going on.

That's right...for suppose f is injective. Let x \in f(AnB). Then f^-1(x)={y} and y is a member of both A and B. Hence, x\in f(A) n f(B) and so f(A n B) \subset f(A) n f (B). To prove the other inclusion, if x \in f(A) n f (B), then y belongs to both A and B. So x \in f(A n B).